Question

What is the arclength of `f(t) = (-2(t+3)^3,(t-2)^2)` on `t in [1,2]`?

Answer 1

`approx 122.006`

Explanation:

We have
`x(t)=-2(t+3)^3`
so
`x'(t)=-6)t+3)^2`
`y(t)=(t-2)^2`
so

`y'(t)=2(t-2)`

so our integral is given by

`int_1^2sqrt((-6(t+3)^2)^2+(2(t-2))^2)dt`
A numerical method gives us `approx 122.006`